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At $27^{o} C N O$ and $C l_{2}$ gases are introduced in a 10 L flask such that their initial pressures are 20 and 16 atm respectively. The flask already contains 24 g magnesium (atomic mass = 24). The amount of magnesium left was 0.2 moles due to the establishment of the following two equilibria:
$2 N O (g) + C l_{2} (g) \underset{}{⇌} 2 N O C l (g)$
$C l_{2} (g) + M g (s) \underset{}{⇌} M g C l_{2} (s); K_{p} = 0.2 a t m^{- 1} .$
The final pressure of $N O C l$ would be:

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7.84 atm

18.06 atm

129.6 atm

64.8 atm

answer is B.

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Detailed Solution

Let x atm $C l_{2}$ reacted in first reaction and y atm $C l_{2}$ reacted in second reaction. Therefore, equilibrium pressures are:

$C l_{2} = 16 - x - y, N O = 20 - 2 x, N O C l = 2 x$

On solving, x = 9 atm and y = 2 atm

Therefore, pressure of NOCl at equilibrium = 18 atm

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